SOLUTION: There were 310 phones and tabs for sale in a shop.All the phones in the shop cost $2484 more than all the tabs. After 1/3 of the phones and 10/11 of the tabs were sold, there were

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Question 802659: There were 310 phones and tabs for sale in a shop.All the phones in the shop cost $2484 more than all the tabs. After 1/3 of the phones and 10/11 of the tabs were sold, there were thrice as many phones as tabs left.If each phone cost $714 more than each tab,how much did each tab cost?
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
= number of phones
= number of tabs
We use the information given to set a system of linear equations to find and .

After 1/3 of the phones were sold, there are left.
After 10/11 of the tabs were sold, there are left.
Since is thrice as many as ,





So -->-->-->
So -->

= price of a tab, in $
so = price of a phone, in $
= how much all the phones originally in the shop cost
$= $2484 more than what all the tabs originally in the shop cost




-->
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